Problem 68 - General Chemistry: Principles and Modern Applications

Question

An \(89.3 \mathrm{mL}\) sample of wet \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(21.3^{\circ} \mathrm{C}\) at a barometric pressure of \(756 \mathrm{mmHg}\) (vapor pressure of water at \(21.3^{\circ} \mathrm{C}=19 \mathrm{mmHg}\) ). (a) What is the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected, in millimeters of mercury? (b) What is the volume percent \(\mathrm{O}_{2}\) in the gas collected? (c) How many grams of \(\mathrm{O}_{2}\) are present in the sample?

Step-by-Step Solution

Answer

(a) The partial pressure of O2 in the gas sample is 737 mmHg. (b) The volume percent of O2 in the gas collected is 100%. (c) The mass of O2 present in the sample is approximately 0.1152 g.

Step 1: Calculate the partial pressure of O2

The gas sample collected is a mixture of O2 and water vapor, hence its total pressure is the sum of the partial pressures of the two gases. According to Dalton's law of partial pressures, the pressures of gases in a mixture add up to the total pressure. Therefore, the pressure of O2 can be calculated by subtracting the vapor pressure of water at the given temperature from the total pressure(Barometric pressure). \nPartial pressure of O2=Total pressure - Vapor pressure of water = 756 mmHg - 19 mmHg = 737 mmHg.

Step 2: Calculation of volume percent of O2

The volume percent of oxygen in the gas mixture can be calculated by dividing the volume of oxygen gas by the total gas volume. Since the gas sample is wet, the total volume is taken as the volume of O2 plus the volume of the water vapor. However, the volume percentages of gases do not depend on the amounts, hence the volume of O2 is equal to total volume. Therefore, Volume% of O2 = (Volume of O2 / Total Volume) x 100 = (89.3 mL / 89.3 mL) x 100 = 100%.

Step 3: Calculate the number of grams of O2

The partial pressure of O2 obtained in step 1 can be used to calculate the amount of oxygen in moles using the Ideal Gas Law (PV=nRT). Here, P is the partial pressure of the gas(737 mmHg or 0.97 atm), V is the volume (89.3 mL or 0.0893 L), R is the ideal gas constant (0.0821 L.atm/K.mol)' and T is the temperature in Kelvin (21.3°C = 294.45 K). First, rearrange the equation to solve for n (number of moles), n = PV / RT. Using these values, n = (0.97 atm x 0.0893 L) / (0.0821 L.atm/K.mol x 294.45 K) = 0.0036 mol. Finally, convert the moles into grams using the molar mass of oxygen (16 g/mol, as oxygen is diatomic, the molar mass is 32 g/mol). Mass = no. of moles x molar mass = 0.0036 mol x 32 g/mol = 0.1152 g.

Key Concepts

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures is a fundamental principle when dealing with gas mixtures. This law states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases. A partial pressure is the pressure that a gas would exert if it occupied the entire volume on its own.
For example, if you collect a sample of gas over water, like in the given exercise, the total pressure recorded (barometric pressure) is the sum of the pressure from the gas (in this case, oxygen) and the vapor pressure from the water. To find the pressure of just the oxygen gas, subtract the water's vapor pressure from the total pressure:

• Total pressure = Pressure of oxygen + Vapor pressure of water
• Thus, Partial pressure of O\(_2\) = Total pressure - Vapor pressure of water.

This simple calculation allows scientists to know the contribution of each gas to the total pressure in a gas mixture.

Ideal Gas Law

The Ideal Gas Law is a critical equation in chemistry to describe the behavior of gases. It is given by the equation \(PV = nRT\), where:

• \(P\) is the pressure of the gas
• \(V\) is the volume of the gas
• \(n\) is the number of moles of gas
• \(R\) is the ideal gas constant, approximately 0.0821 L·atm/mol·K
• \(T\) is the temperature in Kelvin

This law combines several simpler gas laws and allows us to predict how a gas will behave under different conditions of pressure, volume, and temperature.
In this exercise, you use the Ideal Gas Law to find the number of moles of oxygen gas by rearranging it to \(n = \frac{PV}{RT}\). Once you know the moles of gas, you can easily convert that to other measures, like grams, using the molar mass of the gas.

Vapor Pressure

Vapor pressure is a crucial concept in understanding how liquids interact with gases. It is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. As the temperature increases, the vapor pressure of a liquid typically increases as well.
For our gas collection process over water, it is vital to know the vapor pressure of water at the specified temperature to correct for the presence of water vapor in the collected gas. In this exercise, the known vapor pressure of water at 21.3°C is 19 mmHg.
Recognizing the vapor pressure helps us in accurately finding the pressure of the gas of interest—oxygen, in this case—by using Dalton's Law. Understanding vapor pressure is significant not only in laboratory settings but also in various industries like food preservation and meteorology.

Partial Pressure Calculation

Calculating the partial pressure involves knowing both the total pressure and the vapor pressure of any liquid component present. This calculation helps pinpoint the portion of the total pressure that is due to a specific gas in a mixture.

• First, record the total barometric pressure.
• Then, subtract the vapor pressure of water (or other liquids, if applicable) at the given temperature.
• This difference gives you the partial pressure of the main gas—in this instance, oxygen.

The calculated partial pressure is essential for further computations, like determining the amount of gas using the Ideal Gas Law, or analyzing gas mixtures' volumes. It simplifies the complexity of gas mixtures into manageable values that can be further used in chemical reactions and engineering calculations.